Answer:
The roots (zeros) of the function are:
[tex]x=5,\:x=-8[/tex]
Step-by-step explanation:
Given the function
[tex]f\left(x\right)=x^2+3x-40[/tex]
substitute f(x) = 0 to determine the zeros of the function
[tex]0=x^2+3x-40[/tex]
First break the expression x² + 3x - 40 into groups
x² + 3x - 40 = (x² - 5x) + (8x - 40)
Factor out x from x² - 5x: x(x - 5)
Factor out 8 from 8x - 40: 8(x - 5)
Thus, the expression becomes
[tex]0=x\left(x-5\right)+8\left(x-5\right)[/tex]
switch the sides
[tex]x\left(x-5\right)+8\left(x-5\right)=0[/tex]
Factor out common term x - 5
[tex](x - 5) (x + 8) = 0[/tex]
Using the zero factor principle
if ab=0, then a=0 or b=0 (or both a=0 and b=0)
[tex]x-5=0\quad \mathrm{or}\quad \:x+8=0[/tex]
Solve x - 5 = 0
x - 5 = 0
adding 5 to both sides
x - 5 + 5 = 0 + 5
x = 5
solve x + 8 = 0
x + 8 = 0
subtracting 8 from both sides
x + 8 - 8 = 0 - 8
x = -8
Therefore, the roots (zeros) of the function are:
[tex]x=5,\:x=-8[/tex]
